Codeforces Round #262 (Div. 2)解题报告

详见：http://robotcator.logdown.com/posts/221514-codeforces-round-262-div-2

1：A. Vasya and Socks http://codeforces.com/contest/460/problem/A

有n双袜子，每天穿一双然后扔掉。每隔m天买一双新袜子，问最多少天后没有袜子穿。

。

简单思维题：曾经不注重这方面的训练，结果做了比較久。这样的题自己边模拟边想。只是要多考虑trick
```c++
int main(){
int n, m;
long long ans = 0;
scanf("%d%d", &n, &m);
ans = n/m*m;
int tmp = n/m;
int left = tmp + n%m;
while(true){
    ans += left/m*m;
    tmp = left/m;
    left = tmp+left%m;
    if(left < m) break;
}
ans += left;
printf("%I64d\n", ans);
return 0;
}
```
2：B. Little Dima and Equation http://codeforces.com/contest/460/problem/B
题意：
```mathjax
   求满足方程 x = b*S(x)^{a}+c方程解 \\
   0 \lt x \lt 10^{9}   ,\ \ 1 \le a \le 5 ,\ \ 1 \le b \le 10000 ,\ \ -10000 \le c \le 10000\\
   s(x) 为x每位数的和
```
题解：假设简单枚举x肯定超时，可是我们能够换个角度，枚举S(x)。这样就简单多了。由于依据右边就能够算出x..
```c++

int get_sum(long long x){
int ans = 0;
while(x){
    ans += x%10;
    x /= 10;
}
return ans;
}
int main(){
int a, b, c;
scanf("%d%d%d", &a, &b, &c);
long long ans[maxn];
int num = 0;
for(int i = 1; i <= 81; i ++){
    long long tmp = 1;
    for(int j = 1; j <= a; j ++) tmp *= i;
    long long temp = b*tmp + c;
    if(temp > 1e9) continue;
    if(get_sum(temp) == i) {
      ans[num++] = temp;
    }
}
printf("%d\n", num);
if(num > 0) {
    for(int i = 0; i < num; i ++)
      printf("%I64d ", ans[i]);
    printf("\n");
}
return 0;
}
```
3: C. Present http://codeforces.com/contest/460/problem/C
题意：有n朵花，给出初始高度。然后能够浇m次水，每次浇水能够浇连续的w朵，每次浇水后花都会长高1个单位。

问最后最矮的那朵花最大值为多少。
题解：刚開始想，首先要贪心选出最矮的花和其相邻的花浇水。然后是a_(i) 到a_(i+w-1)加一个单位。当时没想到什么好的办法。于是我就想用线段树维护区间最小值，每次找出最小值的下界。然后往右w多花都浇水。刚開始在求下界时想了好久，只是联系一维情况还是写出来了。
```c++
long long a[maxn];
long long mm[4*maxn];
int setv[4*maxn];

void build(int root, int l, int r){
int lc = 2*root, rc = 2*root+1;
if(l < r){
    int mid = (l+r)/2;
    build(2*root, l, mid);
    build(2*root+1, mid+1, r);
    mm[root] = min(mm[lc], mm[rc]);
}else{
    mm[root] = a[l];
}
}

void pushdown(int root, int l, int r){
int lc = 2*root, rc = 2*root+1;
if(setv[root] > 0){
    setv[lc] += setv[root];
    setv[rc] += setv[root];
    mm[lc] += setv[root];
    mm[rc] += setv[root];
    setv[root] = 0;
}
}

void pushup(int root, int l, int r){
int lc = 2*root, rc = 2*root+1;
mm[root] = min(mm[lc], mm[rc]);
}

void modify(int root, int l, int r, int x, int y, int s){
if(x <= l && r <= y){
    mm[root] += s;
    setv[root] += s;
}else{
    pushdown(root, l, r);
    int mid = (l+r)/2;
    if(x <= mid) modify(2*root, l, mid, x, y, s);
    if(y > mid) modify(2*root+1, mid+1, r, x, y, s);
    pushup(root, l, r);
}
}

int minn;
void query(int root, int l, int r, int z){
int lc = 2*root, rc = 2*root+1 , mid = (l+r)/2;
if(l == r){
    if(l < minn) minn = l;
}else{
    pushdown(root, l, r);
    if(mm[lc] <= z) query(lc, l, mid, z);
    else query(rc, mid+1, r, z);
    pushup(root, l, r);
}
}

void print(int root, int l, int r){
    printf("%d %d\n", root, mm[root]);
    if(l < r){
      int mid = (l+r)/2;
      print(2*root, l, mid);
      print(2*root+1, mid+1, r);
    }
}

int main(){
int w, n, m;
scanf("%d%d%d", &n, &m, &w);
for(int i = 1; i <= n; i ++)
    scanf("%d", &a[i]);
memset(setv, 0, sizeof(setv));
build(1, 1, n);
// print(1, 1, n);
for(int i = 1; i <= m; i ++){
    minn = inf;
    query(1, 1, n, mm[1]);
    //cout << minn << endl;
    if(n-minn+1 < w) modify(1, 1, n, n-w+1, n, 1);
    else modify(1, 1, n, minn, minn+w-1, 1);
}
printf("%I64d\n", mm[1]);
return 0;
}
```
昨晚上面两题时间还有15分钟左右，后两题没想出什么好办法。